Difference between revisions of "1967 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
− | <math>\fbox{C}</math> | + | If the candles both have length <math>\ell</math>, then the candle that burns in <math>3</math> hours has a stub of <math>\ell</math> at <math>0</math> minutes, and a stub of <math>0</math> at <math>180</math> minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle <math>t</math> minutes after being lit is <math>f(t) = \frac{\ell}{180}(180 - t)</math>, since <math>f(0) = \ell</math> and <math>f(180) = 0</math>. |
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+ | Similarly, for the second candle that burns out in <math>240</math> minutes, <math>g(t) = \frac{\ell}{240}(240 - t)</math> | ||
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+ | Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub. The problem asks for when <math>g(t) = 2f(t)</math>. Solving this equation gives: | ||
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+ | <math>\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)</math> | ||
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+ | <math>240 - t = \frac{480}{180}(180 - t)</math> | ||
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+ | <math>240 - t = 480 - \frac{480}{180}t</math> | ||
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+ | <math>\frac{8}{3} t - t = 480 - 240</math> | ||
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+ | <math>t = \frac{3}{5} \cdot 240</math> | ||
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+ | <math>t = 144</math> | ||
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+ | So, the second candle will have a stub twice as big as the first candle <math>144</math> minutes after they are both lit. If we want this to happen at <math>4</math> PM, the candles have to be lit <math>144</math> minutes earlier, or <math>2</math> hours and <math>24</math> minutes earlier. This is at <math>\text{1:36 PM}</math>, which is option <math>\fbox{C}</math> | ||
== See also == | == See also == |
Latest revision as of 13:27, 20 July 2019
Problem
Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in hours and the other in hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?
Solution
If the candles both have length , then the candle that burns in hours has a stub of at minutes, and a stub of at minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle minutes after being lit is , since and .
Similarly, for the second candle that burns out in minutes,
Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub. The problem asks for when . Solving this equation gives:
So, the second candle will have a stub twice as big as the first candle minutes after they are both lit. If we want this to happen at PM, the candles have to be lit minutes earlier, or hours and minutes earlier. This is at , which is option
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.